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Math Help - sequences and series

  1. #1
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    sequences and series

    0.9999...=∞(0.9)(0.1)^n-1= 0.9/1-0.1=1,

    n=1

    using the geometric series. By the same method, compute the exact
    value of the following periodic in nite decimal expression as a rational number: 0:89898989
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  2. #2
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    Re: sequences and series

    Quote Originally Posted by Dfaulk044 View Post
    0.9999...=∞(0.9)(0.1)^n-1= 0.9/1-0.1=1,

    n=1
    using the geometric series. By the same method, compute the exact
    value of the following periodic in nite decimal expression as a rational number: 0:89898989
    That is a very very poor way to write or think about it.


    First write {\rm{0:89}}\overline {{\rm{89}}}  = \frac{{89}}{{{{10}^2}}} + \frac{{89}}{{{{10}^4}}} + \frac{{89}}{{{{10}^6}}} +  \cdots  = \sum\limits_{k = 1}^\infty  {\frac{{89}}{{{{10}^{2K}}}}}

    Now your first term is \frac{{89}}{{{{10}^2}}} and the common ratio is \frac{{1}}{{{{10}^2}}}
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  3. #3
    MHF Contributor ebaines's Avatar
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    Re: sequences and series

    First let me rewrite what I think you tried to convey:

     0.999 \overline 9... = \sum _{N=1} ^\infty (0.9)(0.1)^{N-1} = \frac {0.9}{1-0.1} = 1.

    If you do long division of 1-0.1 into 0.9 you get the infinite series .9 + .01 + .009 + ....; Hence 0.99999... = 1.

    For the number 0.898989.. you have:

     0.8989 \overline{89}.. = \sum _{N=1} ^\infty (0.89)(0.01)^{N-1} = \frac {0.89}{1-0.01} = 89/99.

    To show this do the long division of 1 - .01 into 0.89.
    Last edited by ebaines; November 19th 2013 at 11:45 AM.
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